题目：

The  between two integers is the number of positions at which the corresponding bits are different.

Given two integers x and y, calculate the Hamming distance.

Note:

0 ≤ x, y < 231.

Example:

Input: x = 1, y = 4Output: 2Explanation:1   (0 0 0 1)4   (0 1 0 0)       ↑   ↑The above arrows point to positions where the corresponding bits are different.

链接：

3/27/2017

1 public class Solution { 2     public int hammingDistance(int x, int y) { 3         int z = x ^ y; 4         int c = 1, count = 0; 5         while (z != 0) { 6             if ((z & c) != 0) count++; 7             z >>= 1; 8         } 9         return count;10     }11 }

其他人作法：

1. 使用自带函数

1 public class Solution {2     public int hammingDistance(int x, int y) {3         return Integer.bitCount(x ^ y);4     }5 }

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